# Indefinite integral

• Jul 26th 2010, 09:05 AM
cozza
Indefinite integral
Find the following indefinite integral, identifying any rules of calculus that you use:

$\displaystyle \int1/\sqrt{(1-x^2)*\arcsin(x)}$

I think I need to use integration by substitution, so have rearranged to give:

$\displaystyle \int(1-x^2)^{\-1/2} * (sin(x))^{1/2}$

I am not quite sure what to do next. The formula I have for integration by substitution is:

$\displaystyle \int{f(g(x))g'(x)dx} = \int{f(u)du}$, where u = g(x)

Would I then get

f(x) = $\displaystyle (1-x^2)^{\-1/2}$

and

g(x) = $\displaystyle (sin(x))^{1/2}$, so u = -cosx

or do I need to integrate both parts first as they are both composite integrals and then use the above formula on the answers I get? (Headbang)

Thanks for any help in advence
• Jul 26th 2010, 09:21 AM
skeeter
Quote:

Originally Posted by cozza
Find the following indefinite integral, identifying any rules of calculus that you use:

$\displaystyle \int1/\sqrt{(1-x^2)*\arcsin(x)}$

$\displaystyle \displaystyle \int \frac{1}{\sqrt{\arcsin{x}}} \cdot \frac{1}{\sqrt{1-x^2}} \, dx$

$\displaystyle u = \arcsin{x}$

$\displaystyle \displaystyle du = \frac{1}{\sqrt{1-x^2}} \, dx$

substitute ...

$\displaystyle \displaystyle \int \frac{1}{\sqrt{u}} \, du$

finish it
• Jul 26th 2010, 09:33 AM
bondesan
I'll give you a huge hint. You will have to use the u-substitution, this identity:

$\displaystyle \displaystyle{\frac{d}{dx} arcsin(x) = \frac{1}{\sqrt{1-x^2}} = (1-x^2)^{-1/2}}$

and the integral rewritten as: $\displaystyle \displaystyle{\int\frac{1}{\sqrt{(1-x^2)\cdot arcsin(x)}}dx=\int (1-x^2)^{-1/2}\cdot(arcsin(x))^{-1/2}dx}$

This should work.
• Jul 27th 2010, 09:11 AM
cozza
Hi, thanks for the help. Can I just check that the answer I have is right:

$\displaystyle 1/(\sqrt{1-x^2})+c$
• Jul 27th 2010, 09:19 AM
Also sprach Zarathustra

2sqrt(arcsinx) + c
• Jul 27th 2010, 09:26 AM
cozza
Quote:

Originally Posted by Also sprach Zarathustra

2sqrt(arcsinx) + c

Quote:

Originally Posted by cozza
Hi, thanks for the help. Can I just check that the answer I have is right:

$\displaystyle 1/(\sqrt{1-x^2})+c$

On rechecking my answer I don't think the square root should be in there. I think (hope) it is:

$\displaystyle 1/(1-x^2)$ ?

Where does the 2 come from? Have I missed something?
• Jul 28th 2010, 09:30 AM
skeeter
Quote:

Originally Posted by cozza
On rechecking my answer I don't think the square root should be in there. I think (hope) it is:

$\displaystyle 1/(1-x^2)$ ?

Where does the 2 come from? Have I missed something?

yes, you have.

as stated earlier ...

$\displaystyle \displaystyle \int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u} + C$

back substitute $\displaystyle u = \arcsin{x}$ ...

$\displaystyle 2\sqrt{\arcsin{x}} + C$